
The 6th term of a G.P. is 32 and its 8th term is 128, then the common ratio of the G.P. is
A. – 1
B. 2
C. 4
D. – 4
Answer
163.8k+ views
HINT:
The relation \[a{r^{n - 1}}\] is the general formula for the nth term in a geometric series. Here, ‘a’ is the first term of the geometric series, and ‘r’ is the series' common ratio. We can obtain the common ratio by dividing a phrase by its preceding term. We may determine the desired term by replacing values for a, r, and n.
Formula ues:
To calculate Geometric progression, we use the formula
\[a{r^{n - 1}}\]
Where: \[a\] is the first term and \[r\] is the common ratio.
Complete step-by-step solution
We have been given that the \[{6^{th}}\] term of the G.P is \[32\] and its \[{8^{th}}\] term is \[128\]
Here the \[{6^{th}}\] term is \[32\]
So, \[{a_6} = 32\]
The \[{8^{th}}\] term is \[128\]
So, \[{a_8} = 128\]
Now, we have first time and eighth term, we can substitute those values in the formula and find the desired term.
We know that in G.P, the consecutive terms differ by a ratio of \[r\]
Now, we have to calculate the value of \[r\] that is, common ratio for the given G.P
\[{a_6} = {\rm{ a}} \cdot {r^{n - 1}}\]
As, we know that the sixth term is \[32\] we obtain,
\[32 = {\rm{ a}} \cdot {\rm{ }}{r^5} \ldots \ldots (1)\]
\[{a_8} = {\rm{ a}} \cdot {r^7}\]
As, we know that the sixth term is \[128\] we obtain,
\[128 = {\rm{ a}} \cdot {r^7} \ldots \ldots (2)\]
Now, we have to divide equation (2) by equation (1), we have
\[\frac{{a{r^7}}}{{a{r^5}}} = \frac{{128}}{{32}}\]
Let’s cancel the similar terms, we get
\[\frac{{{r^7}}}{{{r^5}}} = 4\]
As base is same in the above equation, powers should be subtracted, so we get
\[{r^{(7 - 5)}} = 4\]
Now, we have to subtract the powers, we obtain
\[{r^2} = 4\]
Let’s square on both sides, we get
\[r = 2\]
Therefore, the the common ratio of the G.P. is \[r = 2\]
Hence, the option B is correct.
NOTE:
Exponentiation can be expressed by trial and error. Students must understand the rule of indices in order to answer such situations. A common ratio exists between terms in a G.P. That is why the power of r increases as the words increase.
The relation \[a{r^{n - 1}}\] is the general formula for the nth term in a geometric series. Here, ‘a’ is the first term of the geometric series, and ‘r’ is the series' common ratio. We can obtain the common ratio by dividing a phrase by its preceding term. We may determine the desired term by replacing values for a, r, and n.
Formula ues:
To calculate Geometric progression, we use the formula
\[a{r^{n - 1}}\]
Where: \[a\] is the first term and \[r\] is the common ratio.
Complete step-by-step solution
We have been given that the \[{6^{th}}\] term of the G.P is \[32\] and its \[{8^{th}}\] term is \[128\]
Here the \[{6^{th}}\] term is \[32\]
So, \[{a_6} = 32\]
The \[{8^{th}}\] term is \[128\]
So, \[{a_8} = 128\]
Now, we have first time and eighth term, we can substitute those values in the formula and find the desired term.
We know that in G.P, the consecutive terms differ by a ratio of \[r\]
Now, we have to calculate the value of \[r\] that is, common ratio for the given G.P
\[{a_6} = {\rm{ a}} \cdot {r^{n - 1}}\]
As, we know that the sixth term is \[32\] we obtain,
\[32 = {\rm{ a}} \cdot {\rm{ }}{r^5} \ldots \ldots (1)\]
\[{a_8} = {\rm{ a}} \cdot {r^7}\]
As, we know that the sixth term is \[128\] we obtain,
\[128 = {\rm{ a}} \cdot {r^7} \ldots \ldots (2)\]
Now, we have to divide equation (2) by equation (1), we have
\[\frac{{a{r^7}}}{{a{r^5}}} = \frac{{128}}{{32}}\]
Let’s cancel the similar terms, we get
\[\frac{{{r^7}}}{{{r^5}}} = 4\]
As base is same in the above equation, powers should be subtracted, so we get
\[{r^{(7 - 5)}} = 4\]
Now, we have to subtract the powers, we obtain
\[{r^2} = 4\]
Let’s square on both sides, we get
\[r = 2\]
Therefore, the the common ratio of the G.P. is \[r = 2\]
Hence, the option B is correct.
NOTE:
Exponentiation can be expressed by trial and error. Students must understand the rule of indices in order to answer such situations. A common ratio exists between terms in a G.P. That is why the power of r increases as the words increase.
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